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Thread: Find all possible x's for cos(x) >= -1/2

  1. #1
    Mar 2010

    Find all possible x's for cos(x) >= -1/2

    Find all x's for this equation.

    $\displaystyle \cos x \geq - \frac{1}{2}$

    When I look at the cosine graph, this looks like the answer

    $\displaystyle (-\frac{2\pi}{3} + 2\pi k) \leq x \leq (\frac{2\pi}{3} + 2\pi k) $, k is a whole number

    but I am not sure. The answer just seems a bit strange. Can someone please tell me if this is the correct way to answer this question?

    (Is there a way I can post the cosine graph in this thread?)

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  2. #2
    Junior Member
    Oct 2010
    The unit circle is a good mental tool for this.

    cos(x) = -1/2 --> x = 2pi/3

    Now, you know that in the interval 0 to 2pi the cosine-function is negative in the second and third quadrants. That is,
    cos(2pi/3) = cos(-2pi/3)

    The inequality thus holds true for the interval
    -2pi/3 < x < 2pi/3
    just like you think. Draw a unit circle and plot the values above.

    Not sure how used you are to working with trigs, give a shout if you need more clarification.
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  3. #3
    MHF Contributor

    Apr 2005
    A simple variation on the previous post. $\displaystyle cos(2\pi/3)= -1/2$ and cosine is decreasing so one part of the set is $\displaystyle [0, 2\pi/3$. But cosine is an even function: cos(-x)= cos(x) so $\displaystyle [-2\pi/3, 0]$ will also work- together, $\displaystyle [-2\pi/3, 2\pi/3]$. But, finally, cosine has period $\displaystyle 2\pi$: $\displaystyle cos(x)\ge -1/2$ for all x in $\displaystyle [-2\pi/3+ 2k\pi, 2\pi/3+ 2k\pi]$ for k any integer, exactly what you have!
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